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How much kinetic energy is needed for the cart to reach the top of the hill?
You are talking about conversion of kinetic energy to potential energy.
PE = mgh = 3kg x 9.81m/s^2 x 22m = 647.46(m^2/s^2)-kg
KE = (1/2)mv^2 = 1.5kg x v^2
KE = PE
1.5kg x v^2 = 647.46(m^2/s^2)-kg The rule is you can do any valid operation on both sides of an equation and it will still be equal. Divide by 1.5kg.
v^2 = 431.64(m^2/s^2) Square root.
v = 20.776m/s <-- ANSWER